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If ABCD is a square, what are the coordinates of C?
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Updated on: 23 Mar 2014, 05:41
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If ABCD is a square, what are the coordinates of C? A. \(\sqrt{3} , \sqrt{3}\) B. \(\sqrt{3}, 1 + \sqrt{3}\) C. \(2 \sqrt{3} , \sqrt{3}\) D. \(1 + \sqrt{3} , \sqrt{3}\) E. \(\sqrt{3} , 2 \sqrt{3}\) Guys  I got the right answer. Please see my attachment in the 2nd file. I have a question though. In my 2nd attachment (Answer.pdf) why can't angle C be 60 degrees and D be 30 degrees?
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Originally posted by enigma123 on 22 Mar 2014, 15:00.
Last edited by Bunuel on 23 Mar 2014, 05:41, edited 1 time in total.
Added the diagram.




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Re: If ABCD is a square, what are the coordinates of C?
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28 Jul 2014, 20:53
GeorgeA023 wrote: If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond. Please refer the screenshot It is a thumb rule; if a right triangle has angles 30  60  90; sides correspondingly opposite would be \(1  \sqrt{3}  2\)
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Re: If ABCD is a square, what are the coordinates of C?
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23 Mar 2014, 05:44
enigma123 wrote: If ABCD is a square, what are the coordinates of C? A. \(\sqrt{3} , \sqrt{3}\) B. \(\sqrt{3}, 1 + \sqrt{3}\) C. \(2 \sqrt{3} , \sqrt{3}\) D. \(1 + \sqrt{3} , \sqrt{3}\) E. \(\sqrt{3} , 2 \sqrt{3}\) Guys  I got the right answer. Please see my attachment in the 2nd file. I have a question though. In my 2nd attachment (Answer.pdf) why can't angle C be 60 degrees and D be 30 degrees? That's because the straight line is 180 degrees (30+90+60) and not 150 degrees (30+90+30).
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Re: If ABCD is a square, what are the coordinates of C?
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24 Mar 2014, 20:17
In this example, all the right triangles formed would be 30  60  90 as per diagram below;
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If ABCD is a square, what are the coordinates of C?
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28 Jul 2014, 17:39
If Side =2 then the diagonal equals \(2\sqrt{2}\)
The diagonal equals the distance between the 2 points. So I assume we can use the distance between 2 points formula.
D=\(\sqrt{(x2x1)^2+(y2y1)^2}\) \(D^2\)=\((x2x1)^2+(y2y1)^2\)
\(2\sqrt{2}^2\)=\((x20)^2+(y21)^2\)
\(2\sqrt{2}^2\)=\((1+\sqrt{3})^2+(\sqrt{3}1)^2\)
8=\((1+\sqrt{3})^2+(\sqrt{3}1)^2\)
8=4+2
???
Can anyone help and tell me where I went wrong.
I thought I could apply the distance between 2 points formula because we knew the distance, and points X1 and Y1.



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Re: If ABCD is a square, what are the coordinates of C?
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28 Jul 2014, 20:26
If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond.



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Re: If ABCD is a square, what are the coordinates of C?
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28 Jul 2014, 20:55
GeorgeA023 wrote: If Side =2 then the diagonal equals \(2\sqrt{2}\)
The diagonal equals the distance between the 2 points. So I assume we can use the distance between 2 points formula.
D=\(\sqrt{(x2x1)^2+(y2y1)^2}\) \(D^2\)=\((x2x1)^2+(y2y1)^2\)
\(2\sqrt{2}^2\)=\((x20)^2+(y21)^2\)
\(2\sqrt{2}^2\)=\((1+\sqrt{3})^2+(\sqrt{3}1)^2\)
8=\((1+\sqrt{3})^2+(\sqrt{3}1)^2\)
8=4+2
???
Can anyone help and tell me where I went wrong.
I thought I could apply the distance between 2 points formula because we knew the distance, and points X1 and Y1. You can. Everything is correct except the last calculation. 8=\((1+\sqrt{3})^2+(\sqrt{3}1)^2\) \(8= 1^2 + \sqrt{3}^2 + 2*1*\sqrt{3} + \sqrt{3}^2 + 1^2  2*\sqrt{3}*1\) \(8 = 1 + 3 + 3 + 1 = 8\)
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Re: If ABCD is a square, what are the coordinates of C?
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28 Jul 2014, 21:11
enigma123 wrote: Attachment: squarecoordinates_figure.PNG If ABCD is a square, what are the coordinates of C? A. \(\sqrt{3} , \sqrt{3}\) B. \(\sqrt{3}, 1 + \sqrt{3}\) C. \(2 \sqrt{3} , \sqrt{3}\) D. \(1 + \sqrt{3} , \sqrt{3}\) E. \(\sqrt{3} , 2 \sqrt{3}\) There are many other ways of solving this question: One method is to note that the sides of 306090 triangle are in the ratio 1:\(\sqrt{3}\):2. Since OA has length 1, OD has length \(\sqrt{3}\). Because the square is tilted, x coordinate of C will be more than x coordinate of D. So options (A), (B) and (E) are out. Also, the square is slightly titled so the x coordinate of C will not be twice of x coordinate of D. Hence option (C) is not possible either. Answer must be (D). Another method is graphing. It is discussed in these two posts: http://www.veritasprep.com/blog/2013/04 ... ryparti/http://www.veritasprep.com/blog/2013/04 ... ypartii/We see that coordinates of D are (\(\sqrt{3}\), 0) because of the 306090 triangle. Now imagine that the x axis is turned 90 degrees at D. So the y coordinate of C will be \(\sqrt{3}\). Also since OA is of length 1, when you rotate it by 90 degrees, the x coordinate of C is 1 more than the x coordinate of C i.e. it is \(\sqrt{3} + 1\).
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Re: If ABCD is a square, what are the coordinates of C?
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29 Jul 2014, 09:42
Karishma,
Thank you so much. This is huge help. I can not believe I missed that.



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Re: If ABCD is a square, what are the coordinates of C?
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06 Aug 2014, 04:53
Is this a special pythagorean triangle like the ones with sides 3,4,5 and 5,12,13?



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Re: If ABCD is a square, what are the coordinates of C?
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06 Aug 2014, 22:41
unceldolan wrote: Is this a special pythagorean triangle like the ones with sides 3,4,5 and 5,12,13? Pythagorean triplets give you the ratio of sides where all three lengths of sides are integers  the most common one is 3, 4, 5. This is the ratio of sides in case the right triangles are special triangles: Triangle with angles 306090 (occur often)  Here ratio of sides is 1: root 3: 2 (Not a pythagorean triplet since all sides are not integers) Isosceles right triangle i.e. angles are 454590  Here ratio of sides is 1:1:root 2 (Not a pythagorean triplet since all sides are not integers)
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Re: If ABCD is a square, what are the coordinates of C?
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18 Jul 2015, 04:18
PareshGmat wrote: GeorgeA023 wrote: If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond. Please refer the screenshot It is a thumb rule; if a right triangle has angles 30  60  90; sides correspondingly opposite would be \(1  \sqrt{3}  2\) Can anyone Pls. explain , how did we get 1 for one of the sides of the blue triangle. I solved till that point and could not proceed.



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Re: If ABCD is a square, what are the coordinates of C?
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18 Jul 2015, 04:40
robinpallickal wrote: Can anyone Pls. explain , how did we get 1 for one of the sides of the blue triangle. I solved till that point and could not proceed.
As mentioned above, after you get the angles as 306090 in the blue triangle, you need to remember that the sides for a 306090 triangle are in the ratio \(1:\sqrt{3}:2\) with '1' side opposite to 30 degree angle. Thus you get 1 for that length. Another way is by trigonometry: in the blue triangle, you know that the hypotenuse =2 Thus in this triangle, cos (60) = base / hypotenuse > 1/2 = base / 2 > base = 1 (as cos(60) = 0.5)



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Re: If ABCD is a square, what are the coordinates of C?
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18 Jul 2015, 05:31
enigma123 wrote: Attachment: squarecoordinates_figure.PNG If ABCD is a square, what are the coordinates of C? A. \(\sqrt{3} , \sqrt{3}\) B. \(\sqrt{3}, 1 + \sqrt{3}\) C. \(2 \sqrt{3} , \sqrt{3}\) D. \(1 + \sqrt{3} , \sqrt{3}\) E. \(\sqrt{3} , 2 \sqrt{3}\) Guys  I got the right answer. Please see my attachment in the 2nd file. I have a question though. In my 2nd attachment (Answer.pdf) why can't angle C be 60 degrees and D be 30 degrees? An explanation using property of slopes! Product of the slopes of Perpendicular lines = 1 \(m_1*m_2 = 1\) Since AO = 1 therefore OD = √3 [306090 Triangle ratio] Let, Slope of Line AD, \(m_1 = 1/√3\) i.e. Slope of Line DC, \(m_2 = √3\) in Triangle i.e. Ratio of YCoordinate of C to Horizontal Distance of CD on XAxis = √3/1 i.e. x Coordinate of C = OD+1= √3+1 i.e. Coordinates of C = ((√3+1),√3) i.e. Option D
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Re: If ABCD is a square, what are the coordinates of C?
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22 Nov 2015, 11:36
since we have 306090 triangle, we can deduct that the sides of the small triangle with the base on the x axis is 1sqrt(3)2 which makes the length of a side of the square = 2. since all the sides are equal, we can draw a line from point c to the xaxis to form a 306090 triangle. since we have a similar triangle with the first one, we see that the hypotenuse is the same = 2. which makes that the point on y axis is sqrt(3)
now, only points A, C, and D have y coordinate of point c sqrt(3). B and E are thus eliminated. now, from the origin to the point D, we have a distance of sqrt(3). The distance from point D to the line perpendicular to the xaxis which form a 306090 triangle is 1. Thus, the xcoordinate of the point C must be 1+sqrt(3)
Eliminate A, and C.
D is the answer.



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Re: If ABCD is a square, what are the coordinates of C?
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11 Apr 2016, 21:11
PareshGmat wrote: GeorgeA023 wrote: If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond. Please refer the screenshot It is a thumb rule; if a right triangle has angles 30  60  90; sides correspondingly opposite would be \(1  \sqrt{3}  2\) For Red Triangle cos 60 =adjancent side/hyp 1/2 =1/h h=2 cos30=\sqrt{3}/2=x/2 x axis length = \sqrt{3} now the blue side cos 60=1/2=x/2 x=1 so total lenght of the x axis is 1+\sqrt{3} only D is the suitable option



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